How do you find the end behavior and state the possible number of x intercepts and the value of the y intercept given $y = x^{4} + 2 x^{2} + 1$ $y=x^4+2x^2+1$ ?

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Answer 1 · 2017-01-01T16:49:20

$y_{intercept} = 1$ $y_("intercept")=1$

No x-intercept.

Set $X = x^{2}$ $X=x^2$ giving:

$y = X^{2} + 2 X + 1$ $y=X^2+2X+1$

$y = {(X + 1)}^{2}$ $y=(X+1)^2$

$\pm (X + 1) = \sqrt{y}$ $+-(X+1)=sqrt(y)$

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$Determine x intercept$ $color(blue)("Determine x intercept")$

Set $y = 0$ $y=0$

$- X - 1 = 0 \Rightarrow x^{2} = - 1$ $-X-1=0 => x^2=-1$

$x = \pm \sqrt{- 1} \to x = \pm i$ $x=+-sqrt(-1)" "->" "x=+-i$

$Thus there is no x_{intercept}$ $color(green)("Thus there is no "x_("intercept"))$
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$Determine y intercept$ $color(blue)("Determine y intercept")$

$y_{intercept}$ $y_("intercept")$ occurs at $X = 0$ $X=0$

$y = X^{2} + 2 X + 1 \to y = 0^{2} + 2 (0) + 1 = 1$ $y=X^2+2X+1" "->" "y=0^2+2(0)+1 = 1$

$y_{intercept} = 1$ $color(green)(y_("intercept")=1$
Tony B

How do you find the end behavior and state the possible number of x intercepts and the value of the y intercept given y=x^4+2x^2+1y=x4+2x2+1y=x^4+2x^2+1?